MATH 615 LECTURE NOTES, WINTER, 2010 by Mel Hochster. Lecture of January 6, 2010

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1 MATH 615 LECTURE NOTES, WINTER, 2010 by Mel Hochster ZARISKI S MAIN THEOREM, STRUCTURE OF SMOOTH, UNRAMIFIED, AND ÉTALE HOMOMORPHISMS, HENSELIAN RINGS AND HENSELIZATION, ARTIN APPROXIMATION, AND REDUCTION TO CHARACTERISTIC p Lecture of January 6, 2010 Throughout these lectures, unless otherwise indicated, all rings are commutative, associative rings with multiplicative identity and ring homomorphisms are unital, i.e., they are assumed to preserve the identity. If R is a ring, a given R-module M is also assumed to be unital, i.e., 1 m = m for all m M. We shall use N, Z, Q, and R and C to denote the nonnegative integers, the integers, the rational numbers, the real numbers, and the complex numbers, respectively. Our focus is very strongly on Noetherian rings, i.e., rings in which every ideal is finitely generated. Our objective will be to prove results, many of them very deep, that imply that many questions about arbitrary Noetherian rings can be reduced to the case of finitely generated algebras over a field (if the original ring contains a field) or over a discrete valuation ring (DVR), by which we shall always mean a Noetherian discrete valuation domain. Such a domain V is characterized by having just one maximal ideal, which is principal, say pv, and is such that every nonzero element can be written uniquely in the form up n where u is a unit and n N. The formal power series ring K[[x]] in one variable over a field K is an example in which p = x. Another is the ring of p-adic integers for some prime p > 0, in which case the prime used does, in fact, generate the maximal ideal. One can make this sort of reduction in steps as follows. First reduce to the problem to the local case. Then complete, so that one only needs to consider the problem for complete local rings. We shall study Henselian rings and the process of Henselization. We shall give numerous characterizations of Henselian rings. In good cases, the Henselization consists of the elements of the completion algebraic over the original ring. The next step is to approximate the complete ring in the sense of writing it as a direct limit of Henselian rings that are Henselizations of local rings of finitely generated algebras over a field or DVR. But this is done in a good way, where many additional conditions are satisfied. The result needed is referred to as Artin approximation. We are not yet done. Henselizations are constructed as direct limits of localized étale extensions, and so we are led to study étale and other important classes of ring extensions, such as smooth extensions and unramified extensiions. (The étale extensions are the extensions that are both smooth and unramified.) There is a beautiful structure theory for 1

2 2 these classes of extensions. Because étale extensions are finitely generated algebras, one can take the fourth step, which is to replace the Henselian ring by a ring that is finitely generated over a field or DVR. Carrying out these ideas in detail will take up a large portion of these notes. Étale extensions have numerous applications to geometry: they are used to remedy the fact that the implicit function theorem does not hold in the allgebraic context in the same sense that it is does when working with C or analytic functions. As an example, we shall later use the theory of étale extensions to establish a relationship that is not obvious between intersection multiplicities defined algebraically and intersection multiplicities defined quite geometrically. The structure theorems we want to prove depend on an algebraic result known as Zariski s Main Theorem, or ZMT. It has many applications in commutative algebra and algebraic geometry. In our formal treatment, we shall first prove Zariski s Main Theorem, and then define and analyze the structure of smooth, étale, and unramified homomorphisms. We shall discuss Henselization, Artin approximation, and applications in which one reduces questions about arbitrary Noetherian rings to the case of algebras finitely generated over a field or a discrete valuation ring (DVR). Another tool that we introduce provides a method for reducing many questions about finitely generated algebras over a field of characteristic 0 to corresponding questions for finitely generated algebras over a field of characteristic p > 0: in fact to the case where that field is finite! It may be surprising that one can do this: it turns out to be a very powerful technique. I do want to emphasize that the theory we build here shows that the study of finitely generated algebras over a field or DVR is absolutely central to the study of arbitrary Noetherian rings. Before stating Zariski s Main Theorem, we review some facts from commutative algebra that we assume in the sequel. Following the review, we state the algebraic form of the theorem, review some basic algebraic geometry, and then give a geometric version of ZMT. We explain how to deduce the geometric version from the algebraic version, and then go to work on the proof of the algebraic version, which is rather long and difficult. A prime ideal of R is a a proper ideal such that R/P is an integral domain. The (0) ideal is prime if and only if R is an integral domain. The unit ideal is never prime. The set of prime ideals of R, denoted Spec (R) is a topological space in the Zariski topology, which is characterized by the fact that a set of primes is closed if and only if it has the the form V(I) = {P Spec (R) : I P }. I may be any subset of R, but V(I) is unchanged by replacing I by the ideal it generates, so that one may assume that I is an ideal. When I is an ideal, V (I) is unchanged by replacing I by Rad (I) = {r R : for some integer n > 0, r n I}. The closed sets of Spec (R) are in bijective order-reversing correspondence with the radical ideals of R. The closure of the point given by the prime ideal P is V(P ), so that P is a closed point if and only if P is a maximal ideal of R. Note that Spec (R) is not, in general, T 1.

3 If f R, D f denotes Spec (R) V(Rf), the set of prime ideals of R not containing f. The sets D f are a basis for the open sets of the Zariski topology on R. Note that D fg = D f D g. Let h : R S be a ring homomorphism. Then h or Spec (h) denotes the map Spec (S) Spec (R) whose value on Q Spec (S) is the inverse image h 1 (Q) under h. This inverse image is also called the contraction of Q to R. Note that if R S, the contraction of Q to R is simply Q R. We assume some familiarity with categories and functors. Spec is a contavariant functor from the category of commutative rings and ring homomorphisms to the category of topological spaces and continuous maps. (Very briefly, functors assign values to objects and morphisms in a category in such a way that identity maps are preserved, and composition is either preserved or reversed. Functors preserving composition are called covariant, while those reversing composition are called contravariant.) A multiplicative system W in a ring R is a subset that contains 1 is closed under multiplication. The localization of R at W, denoted W 1 R, is an R-algebra in which every element of W becomes invertible. Every R-module M also has a localization at W, denoted W 1 M, which is a W 1 R-module. (W 1 M may be defined as equivalence classes of pairs (m, w) M W where (m, w) is equivalent to (m, w ) if there exists v W such that v(w m wm ) = 0. The equivalent class of (m, w) is denote m/w. W 1 M and an W 1 R-module. Addition and multiplication by scalars are such that (m/w) + (m /w ) = (w m + wm )/(ww ), r(m/w) = (rm)/w, and (r/v)(m/w) = (rm)/((vw). There is an R-linear map M W 1 M that sends m m/1. This map need not be injective. In fact, the kernel consists of all elements m M such that wm = 0 for some w W. These remarks include the case M = R. Note that W 1 R is a ring, and the multiplication satisfies (r/w)(r/w ) = (rr )/(ww ). M W 1 M is injective if and only if no element of W is a zerodivisor on M, i.e., multiplication by every w W gives an injective map M M. Note also that a homomorphism R S can be factored R W 1 R S if and only if the image of W in S consists entirely of units, in which case the factorization is unique. This is referred to as the universal mapping property of localization. The notation M W is used as an alternative to W 1 M, but we will not use this notation in these notes. There is a canonical isomorphism W 1 R R M W 1 M such that (r/w) m (rm)/w and, under the inverse isomorphism, m/w (1/w) m. M W 1 M is a covariant exact functor from R-modules to W 1 R-modules: if f : M N, there is a unique map W 1 M W 1 N such that m/w f(m)/w. If P is a prime ideal of R, W = R P is a multiplicative system (this characterizes which ideals are prime). In this case W 1 R is denoted R P and is called the localization of R at P. Likewise, W 1 M is denoted M P. There is a canonical isomorphism of R P R = MP. If S = W 1 R, there is a bijective homeomorphism between Spec (W 1 R) and {P Spec (R) : W P = } (with the inherited Zariski toipology from Spec (R). The maps send Q Spec (W 1 R) to its contraction to R, and P Spec (R) to its expansion P S to S. (For any homomorphism 3

4 4 R S, if I is an ideal of R its expansion IS is the ideal of S generated by the image of I.) Thus, there is bijection between the primes of W 1 R and the primes of R that do not meet W. If R S is surjective, then S = R/I, where I is the kernel. In this case there is a homeomorphism of Spec (R/I) with V(I) Spec (R), again given by contraction and expansion. When S is an R-algebra and W is a multiplicative system in S, we have a definition for W 1 S: we may think of S as an R-module. This is canonically isomorphic with S-algebra V 1 S obtained by localizing S at the image of W. If P is a prime ideal and W = R P, this gives an identification of R P /P R P with the fraction field of R/P. This field is often denoted κ P, but the notation is ambiguous, since it conceals the dependence on R. If W is a multiplicative system in R and I is an ideal of R, the R-algebras S = W 1 (R/I) and W 1 R/IW 1 R are canonically isomorphic. In this case we may put the facts above together to conclude that Spec (S) Spec (R) gives a homeomorphism of Spec (S) with the set of prime ideals of R that contain I and are disjoint from W (in the inherited Zariski topology from Spec (R). Let h : R S be a ring homomorphism and let f = Spec (h) be the continuous map Spec (S) Spec (R) given by contraction of prime ideals.. (Sometimes f is denote h.) The set-theoretic fiber of f over a prime P is f 1 (P ), i.e., the set of primes Q of S that contract to P. The primes that contract to P are also said to lie over P. By taking I = P and W = R P, this set of primes may be identified with Spec (W 1 )S/P S, for Q lies over P if and only if it contains the image of P, and, hence, P S, and is disjoint from the image of W. The ring W 1 (S/P S) = (W 1 S)/(P W 1 S) = κ P R S is called the scheme-theoretic fiber of R S over P. This point of view enables one to think of the set-theoretic fiber f 1 (P ) as the space of prime ideals of the scheme-theoretic fiber, (R P ) 1 S/P (R P ) 1 S = κ P R S. A prime Q of S that lies over P in R is called isolated in its fiber or isolated in the fiber over P if it is both maximal and minimal among primes lying over P. In particular, if Q is the unique prime lying over P then it is isolated in its fiber. We shall return to this notion soon and explain the use of the word isolated here, but we first want to state Zariski s Main Theorem, which we sometimes abbreviate ZMT. We next want to recall the notion of integral dependence of elements. If R S we say that an element s S is integral over R if it is a root of some monic polynomial with coefficients in R. In other words, s satsifies an equation of the form s n + r n 1 s n r j s j + + r 1 s + r 0 = 0 where n is a positive integer and the r j R. The set of elements of S integral over R is a subring of S containing R called the integral closure of R in S. R is said to be integrally closed in S if the integral closure of R in S is R, i.e., every element of S integral over R is in R. An integral domain is called integrally closed or normal if it is integrally closed in its fraction field. Every unique factorization domain (UFD) is normal.

5 Theorem (Zariski s Main Theorem. Suppose that R R[θ 1,..., θ n ] S are commutative rings and that R is integrally closed in S while S is integral over R[θ 1,..., θ n ]. Let Q be a prime ideal of S that is isolated in its fiber over P Spec (R). Then there exists an element f R P such that the induced homomorphism R f S f is an isomorphism. We want to examine some consequences of this result, including a very important geometric corollary, before we give the proof, which is difficult and lengthy. We next review some basic algebraic geometry. Let K be an algebraically closed field. For simplicity, at this point we shall restrict our attention primarily to closed algebraic sets in some A N K. Let X be such a set. Unless otherwise specified, when we refer to points of X we mean closed points. By a variety we mean a nonempty irreducible closed algebraic set in some A N K. (As a scheme, a variety is reduced and irreducible.) We write K[X] for the coordinate ring of the affine (closed) algebraic set X: it is the ring of regular functions from X to K = A 1 K. It is a finitely generated reduced K-algebra (and, up to isomorphism, all such algebras occur). X is a variety iff K[X] is a domain. Note that the category of (closed) affine algebraic sets over K and regular morphisms is anti-equivalent to the category of finitely generated reduced K-algebras and K-algebra homomorphisms: essentially, these are opposite categories. If X is a variety then the fraction field of K[X] is denoted K(X) and is called the function field of X. Its elements may be regarded as regular functions defined on some nonempty (equivalently, dense) open set in X, where two functions are equivalent if they agree on the intersection of their domains, which will be another dense open set. A morphism g : X Y of varieties is called dominant if its image is dense. This holds if and only if the induced map of K[Y ] K[X] is injective, for that map has kernel containing I if and only if the image of g is contained V (I) Y. A dominant map induces a map of function fields K(Y ) K(X), which is necessarily injective. By definition, the variety Y is normal precisely when K[Y ] is normal, i.e., integrally closed in its field of fractions K(Y ). The following is a corollary of ZMT, and is also referred to as Zariski s Main Theorem. The restriction to affine varieties is not needed and is only made for simplicity. We shall explain how the Corollary is deduced in detail later. Corollary (Zariski s Main Theorem). Let g : X Y, be a morphism of affine varieties as in the preceding discussion. If g is bijective on closed points, Y is normal, and K(X) is separable over K(Y ), then g is an isomorphism. We have not yet proved anything. We first want to discuss why some hypothesis other than having g be bijective on closed points is needed. Let Y be V (y 3 z 2 ) in A 2 K, which may also be described as the set {(λ2, λ 3 ) : λ K}. Then K[Y ] = K[y, z]/(y 3 z 2 ) = K[x 2, x 3 ] K[x]. Let X = A 1 K. The map X Y that sends λ to (λ 2, λ 3 ) is bijective. It corresponds to the map K[Y ] = K[y, z]/(y 3 z 2 ) = K[x 2, x 3 ] K[x] = K[X] that sends the images of y and z to x 2 and x 3, respectively. This is an example of a bijective map of varieties that is not an isomorphism: the problem, in some sense, is that K[Y ] is not normal the element x is in its integral closure. Thus, Y is not normal. Note that the map of varieties cannot be an isomorphism because the corresponding map of K-algebras is not surjective, and therefore is not an isomorphism. 5

6 6 Even when both varieties are normal, or even regular, the separability condition is needed. Let K be an algebraically closed field of prime characteristic p > 0 and let X = Y = A 1 K. Let g : X Y be the map sending λ λp. Since K is algebraically closed it is perfect, and so the map is surjective and therefore bijective. This morphism corresponds to the K-algebra map of rings K[x] K[x] sending x x p, or to the inclusion K[x p ] K[x]. The map of rings is not surjective and so g is not an isomorphism. The induced map of function fields is K(x p ) K(x), which is evidently not separable: that is the problem. We next want to note that ZMT is rather non-trivial even in very special cases: it implies a key lemma that can be used to deduce Hilbert s Nullstellensatz very quickly. Suppose that in the statement of (the ring-theoretic form of) ZMT one assume that R = K L = S where K is an algebraic closed field and S is a field that is finitely generated as a K-algebra. Then ZMT applies: K is integrally closed in L because it is algebraically closed in L. Take P = (0) K and Q = (0) L as the two primes (of course, there are no other primes to choose). Evidently Q is isolated in its fiber, since L has only one prime. Then there exists f K P such that K f = Lf. Since f 0 it is already invertible in K and L, and so we see that K = L. That is, a field finitely generated as an algebra over an algebraically closed field K must be equal to K. This result is sometimes called Zariski s lemma. However, our proof of ZMT will not give a new proof of Hilbert s Nullstellensatz: we make use of Hilbert s Nullstellensatz in the argument. We next want to explain the used of the word isolated in the expression isolated in its fiber. A point x of a topological space X is called isolated if it is both open and closed in X. The fiber, as a topological space, is the Spec of the ring A = κ P R S, and so may be thought of as a topological space. We want to make two observations: (1) If a prime m is an isolated point of Spec (A), then m is both maximal and minimal among the prime ideals of A. (2) If A is Noetherian or, much more generally, if the prime ideal m is finitely generated, then m is an isolated point of Spec (A) if and only if m is both maximal and minimal in Spec (A). To see this, first note that {m} is closed if and only if m is maximal. Now {m} is open if and only if there exists f R such that mr f is the unique prime ideal of R f : this is the condition for D(f) = {m}. Since a prime has arbitrarily small open neighborhoods of the form D(f), the set consisting of just that prime cannot be open unless it is equal to D(f) for some choice of f. But if m = D(f) it must be minimal: any strictly smaller prime would also be in D(f). Finally, suppose that m is both maximal and minimal and that it is finitely generated, say by u 1,..., u h. If we localize at m it becomes the only prime of R m, and so every u j has become nilpotent. This implies that for every j we can choose f j R m and N j such that f j u N j = 0. Let f = f 1 f h. In the ring R f, every generator of m is nilpotent. Since m is maximal, it is the only prime ideal of R f, and thus {m} is open as well as closed. We next want to show how the geometric form of ZMT stated above follows from the algebraic form. We need the following basic facts about the behavior of dominant maps of

7 7 varieties: Lemma. Let g : X Y be a dominant map of algebraic varieties, so that we have an injection of domains K[Y ] K[X]. Then: (a) The transcendence degree of K(X) over K(Y ) is δ = dim (X) dim (Y ). (b) There is a dense open subset U of Y such that for every u U, the dimension of the fiber g 1 (u), thought of as a closed algebraic set in X, is δ = dim (X) dim (Y ). (c) If dim (Y ) = dim (X) then K(X) is a finite algebraic extension of K(Y ). Assume also that K(X) is separable over K(Y ). Then there is a dense open set U Y such that for all u U, the fiber g 1 (u) is a finite set of cardinality d = [K(X) : K(Y )]. Proof. Given any three fields K F G the transcendence degree of G over K is the sum of the transcendence degree of F over K and the transcendence degree of G over F. Part (a) follows from applying this to K K(Y ) K(X) along with the theorem that the dimension of a variety over K is the transcendence degree of its function field over K. To prove part (b), let R = K[Y ] K[X] = S. Then S is a domain finitely generated over the domain R, and by the Noether normalization theorem for domains, we may localize at one nonzero element f R so that S f is a module-finite extension of a polynomial ring over R. The number of variables must be δ, the transcendence degree. Let U be the open set corresponding to D(f) in Y. Thus, after replacing R and S by R f and S f, it suffices to show that if S is a module-finite domain extension of R[x 1,..., x δ ], then all fibers over maximal ideals m of R have dimension δ. Since S/mS is module-finite over (R/m)[x 1,..., x δ ] the dimension is at most δ. Since S has prime ideal Q lying over mr[x 1,..., x δ ] by the lying over theorem, and we have S/mS S/Q, while S/Q is a module-finite extension domain of (R/m)]x 1,..., x δ ], we also have that the dimension is at least δ. It remains to consider part (c). We continue the notations from the proof of (b). The first statement is immediate from (a) and the fact that S = K[X] is finitely generated over K and, hence, over R = K[Y ]. We may localize at f R {0} and so assume that S is module-finite over R. Then K(Y ) R S = K(X). Choose a primitive element θ for K(X) over K(Y ): by multiplying by a suitable nonzero element in R, we may assume that θ is in S. Let G be the minimal monic polynomial of θ over the fraction field of R. By our hypothesis on the field extension, G will be separable over R. By inverting one more element of R {0} we may assume that the coefficients of G are in R. Note that S/R[θ], as an R-module, is torsion. Therefore we may invert yet another element of R and assume without loss of generality that S = R[θ], and then S = R[x]/G. Consider the roots of G in a suitably large extension field of the fraction field of R. The product of the squares of their differences (the discriminant of G) is a symmetric polynomial over Z in the roots of G, and therefore is expressible as a polynomial D over Z in the coefficients of G, which, up to sign, are the elementary symmetric functions of the roots. The discriminant is therefore a nonzero element of R. We localize at the discriminant as well, and so we may assume that it is a unit of R. Note that each localization has the effect of restricting out attention to a smaller dense open subset of Y.

8 8 The points of the fiber over a m, a maximal ideal of R, correspond to the maximal ideals of (R/m)[x]/G, where G is simply the image of G modulo m. But R/m = K and the discriminant of G is simply the image of the discriminant of G (one substitutes the images of the coefficients of G into D), and so is not zero. It follows that the roots of G are mutually distinct, and so the number of points in the finite fiber is precisely the degree of G, which is the same as the degree of G and is equal to [K(Y ) : K(X)]. Lecture of Janauary 8, 2010 Proof of the geometric form of ZMT. We are now ready to deduce the geometric form of ZMT from the algebraic form. The fact that the map g : X Y is bijective implies that it is dominant. Therefore, we may consider R = K[Y ] K[X] = S. We want to prove that R = S. Consider S/R as an R-module. If it is nonzero, then there is a maximal ideal m of R such that (S/R) m 0, and then R m S m, where S m is simply (R m) 1 S. The bijectivity of the map shows that all set-theoretic fibers over closed points consist of exactly one closed point. From part (b) of the preceding Lemma, we must have dim (X) = dim (Y ), and then part (c) shows that the extension is algebraic of degree 1, which means that K(X) = K(Y ): note that we are using separability here. Since R m is normal, it is integrally closed in S m, for S m is contained in the fraction field of R. Since S is finitely generated over R, S m is finitely generated over R m. The fiber S m /ms m contains just one prime ideal, which is isolated in its fiber. Therefore we can choose f R m mr m such that (R m ) f = (S m ) f. But since f is already invertible in R m and S m, this implies that R m = S m, a contradiction. We are now ready to begin the proof of the algebraic form of ZMT. We have that Q Spec (S) is isolated in its fiber over P Spec (R), where R[θ 1,..., θ n ] S is integral, while R is integrally closed in S. We shall used induction on n. If n = 0, then S is integral over R, and since R is integrally closed in S, this implies that R = S, and we are done. We postpone considering the case where n = 1. Instead, we assume this case for the moment, and carry through the inductive step. This will reduce the problem to studying the case where n = 1. For this purpose let T denote the integral closure of R[θ 1,..., θ n 1 ] in S. Let θ = θ n. Note that S is integral over T [θ]. Let q = Q T. We want to show that q is isolated in its fiber over R, which is the fiber over P, since it is clear that q R = Q R = P. Suppose that we can do this. Then we can choose f R P such that R f = T f, by the induction hypothesis. Now Q is evidently isolated in its fiber over q, since q lies over P in R, and this is preserved when we localize at f. Also, S is integral over T [θ], and so S f is integral over T f [θ]. Thus, we can apply the case n = 1 to conclude that there exists γ = T f qt f such that (T f ) γ = (S f ) γ. Now, T f = R f, and it follows that q f = P R f, so that we may write γ = g/f h where g R P. This gives R fg = (R f ) γ = (T f ) γ = (S f ) γ = S fg, as required. It remains to prove that q is isolated in its fiber over P. We need the following result (think of S 0 as R[θ 1,..., θ n ]).

9 Lemma. Let R T S be rings. Let Q be a prime of S lying over P = Q R, and let q = Q T. (a) If Q is minimal in its fiber over T and u T q is such that T u = S u, then q is also minimal in its fiber over P. (b) If S is integral over S 0, R S 0 S with S 0 finitely generated over R, and Q is maximal in its fiber over T, then q is maximal in its fiber over P. (c) If Q is isolated in its fiber, S is integral over a finitely generated R-subalgebra S 0, and there exists u T q such that T u = S u, then q is isolated in its fiber over T. Proof. To prove part (a) note that if q is strictly contained in q and q lies over P, we may expand q to T u = S u and then contract to S to obtain a prime Q of S strictly smaller than Q and lying over P : we have a bijection between primes of T not containing u and prime of S not containing u. For part (b), first replace R, T, and S by their localizations at P. Thus, we may assume that R is local with maximal ideal P. Then R/P T/q S/Q, and we also have R/P S 0 /q S/Q, where q = Q S 0. The fact that S is maximal in its fiber implies that S/Q is a field. Since S/Q is integral over S 0 /q, S 0 /q is also a field, and it is finitely generated over R/P. It follows that S 0 /q is a finite algebraic extension of R/P. Now we get that S/Q is an algebraic extension of R/P. It now follows that T/q is a field, and this means that q is maximal in its fiber. Part (c) is simply the result of combining (a) and (b). We can now use this Lemma to to see that q as defined earlier is isolated in its fiber over P. We have the hypothesis of part (b). The hypothesis of part (a) also holds, because we may apply the case n = 1 to the inclusion T S (Q is obviously isolated in its fiber over q) to obtain u T q such that T f = S f ). The result we need is now immediate from part (c). This completes the reduction to the case where n = 1. 9 Lecture of Janauary 11, 2010 For the remainder of the proof we shall be studying the case where R R[θ] S. R is integrally closed in S, S is integral over R[θ], and Q, a prime of S, is isolated in its fiber over P = Q R. We want to reduce to the case where S is actually module-finite over R[θ]. For this purpose it will suffice to find T module-finite over R[θ] such that R[θ] T S and q = Q T is isolated in its fiber over P. For once we have T as specified, if we know the theorem in the module-finite case we can choose f R P such that R f = T f S f. Then R f is integrally closed in S f and S f is integral over T f = R f together imply that R f = S f. Quite generally, we are free to replace S by any ring T with R[θ] T S such that Q T is minimal in its fiber: by the argument just given, if the desired result holds for T then it holds for S. Note that by part (b) of the Lemma of Janaury 8, we have that q is maximal in its fiber for any choice of T with R[θ] T S. Thus, the problem is to choose T such that q is minimal in its fiber. Thus, we want to find s 0, s 1,..., s h S such that with T = R[θ][s 0, s 1,..., s h ], q = Q T is minimal in its fiber over P. For this purpose we may replace R, S by R P, S P and θ by its image in S P. If we find elements

10 10 s 0 /f 0, s 1 /f 1,..., s h /f h that work here, where the s j S and the f j R P, we may use s 0, s 1,..., s h, since the f j become invertible in any case when we calculate the fiber. Let K = R/P. Let θ be the image of θ in R[θ]/P R[θ] = K[θ]. If θ satisfies a monic polynomial of positive degree over K, then we may take T = R[θ]: the fiber is module-finite over K and, hence, zero-dimensional, and so all primes of the fiber of T over P will be isolated in that case. Therefore, we may assume without loss of generality that θ is transcendental over K. Q lies over a prime of this ring that is maximal in its fiber, by part (b) of the Lemma of Janauary 8, and this prime will be generated by g(θ), where g is a monic polynomial over K. Lift this element to an element γ R[θ]. Then γ Q is nilpotent mod P S, since Q is minimal in its fiber. Hence, we can choose s 0 S Q, p 1,..., p h P, and s 1,..., s h S such that s 0 γ N = h j=1 s jp j. Choose T = R[θ][s 0, s 1,..., s h ]. We claim that q = Q T is minimal in its fiber over P, for a smaller prime q 0 that lies over P will not contain s 0, and so must contain γ. But then q and q 0 contract to the same prime in R[θ], since there is only one prime that contains P and γ, and since T is integral over R[θ] it follows that q = q 0. Once we have replaced S by a module-finite extension of R[θ], we may again replace R by R P and S by S P, and it suffices to see that R P = S P. For if these two are equal, then for each of the finitely many generators of S over R we may choose f j multiplying that generator into R with f j R P. Let f be the product of the f j. Then R f = S f. Henceforth we assume that R = (R, P, K) is quasi-local as well. If A B is module-finite the conductor of B in A is defined as {a A : Ba A}. It is readily checked to be an ideal of A and an ideal of B. It is also easy to see that it is the largest ideal of A that is also an ideal of B, since any element a of such an ideal has the property that Ba A. Throughout the rest of the proof of ZMT, let J be the conductor of S in R[θ]. The remainder of the proof breaks up into two cases: one where J Q, which is easier, and one where J Q. The second case will require two preliminary results. We first discuss the case where J Q. Let u J Q. Then R[θ] u = S u, and by part (a) of the Lemma of January 8, Q R[θ] is isolated in its fiber over P. We may therefore replace S by R[θ]: once we have the result for R[θ], the case of S follows, by the comment at the end of the first paragraph of the preceding page. But then R[θ]/P R[θ] = K[θ] cannot be a polynomial ring in θ over K, for no prime could then be isolated. It follows that there is a polynomial in θ with at least one coefficient not in P whose value is in P R[θ]. Subtracting, we find that there is a polynomial in θ over R that vanishes and has at least one coefficient not in P. Choose such a polynomial of least degree d, and call it r d θ n + r d 1 θ d r 0. By multiplying the polynomial by r d 1 d, we see that r d θ is integral over R and therefore in R. Thus, we may re-write the polynomial as (r d θ + r d 1 )θ d r 0, which has lower degree in θ. Therefore all of its coefficients are in P, and we conclude that r d θ + r d 1 P. If r d is a unit, we find that θ R, as required. If r d P and r d 1 is a unit, we find that r d 1 P S, a contradiction, since Q contains P S. This completes the proof for the case where J Q.

11 In the remaining and most difficult case, where J Q, we shall show that Q cannot be isolated in its fiber. We need two preliminary results. Lemma. If R R[θ] S are domains with S integral over R[θ] and θ is transcendental over R, then no prime ideal of S is isolated in its fiber. Proof. Suppose that Q is isolated in its fiber over P = S R. Let S be the integral closure of S in its fraction field. By the lying over theorem, there exists a prime ideal Q of S lying over Q. Q must also be isolated in its fiber over P : a prime Q 1 comparable to but distinct from it lying over P would yield such a prime lying over P and comparable to but distinct from Q when contracted to S (Q 1 cannot lie over Q: primes lying over the same prime in an integral extension are mutually incomparable). Henceforth we assume that S = S is integrally closed. It then contains an integral closure R of R in the fraction field of R. Then Q will be isolated in its fiber over P = Q R : a comparable prime lying over P will automatically lie over P (note that P R Q R = P ). Therefore we may replace R by R and R[θ] by R [θ], and so assume that R is integrally closed. But now both going up and going down hold between R[θ] and S, since R[θ] is again normal, and it follows that Q R[θ] is also isolated in its fiber over P. We may consequently replace S by R[θ]. But now we can see that no prime is isolated in its fiber, since the fiber over P is the polynomial ring in one variable over a field, and there is no prime that is both maximal and minimal. 11 We need one more preliminary result: Lecture of January 13, 2010 Lemma. If A A[τ] B with B integral over A[τ] and A integrally closed in B, and there is a monic polynomial F with coefficients in A such that F (τ)b A[τ], then B = A[τ]. Proof. Let b B be arbitrary. Then F (τ)b = G(τ) for some polynomial G with coefficients in A, since F (τ)b A[τ]. By the division algorithm for monic polynomials we can write G = QF + H, where H is either 0 or of degree smaller than that of F, and where Q and H are polynomials with coefficients in A. Let c = b Q(τ). It will suffice to show that c A, and for this it will suffice to show that c is integral over A. Now F (τ)b = F (τ)(q(τ)+c), but F (τ)b = G(τ) = Q(τ)F (τ) + H(τ) as well, and so F (τ)c = H(τ). Since deg(h) < deg(f ), this implies that τ/1 B c is integral over the ring A c, where A denotes the image of A in B c. Since B is integral over A[τ], we have that B c is integral over A c, and, in particular, c/1 B c is integral over A c. If we write down an equation of integral dependence in which the coefficients have common denominator c M, we get ( c 1 ) d + a d 1 c M and, multiplying by c M, we have that ( c ) d 1 a c M = 0 c M+d + a d 1 c d a 0

12 12 has image 0 in B c, and so is killed by a power of c. This shows that c is integral over A, and so is in A, as required. Proof of Zariski s Main Theorem: the finale. We shall now show that if J, the conductor of S into R[t], is contained in Q that Q is not isolated in its fiber, which will complete the proof of the theorem. Recall that R is integrally closed in S and that S is module-finite over R[θ]. If J Q we can choose a minimal prime q of J in S such that q Q. Let p = q R. Let t denote the image of θ in the ring S/q. Now, R/p (R/p)[t] S/q, and S/q is integral over (R/p)[t]. The fact that Q is isolated in its fiber over P implies that Q/q is isolated in its fiber over P/p R/p. By the final Lemma in the Lecture Notes from January 11, this means that t cannot be transcendental over R/p. We complete the proof by obtaining a contradiction. If t is algebraic over R/p, localize at p and consider the image τ of θ in S p. We then obtain a monic polynomial F with coefficients in R p such that F (τ) qs p. Since q is a minimal prime of J, there exist a positive integer N and w S q such that w ( F (τ) ) N JSp. Replacing F by F N, we have F monic over R p such that wf (τ) JS p. We now apply the Lemma above with A = R p and Note that since F (τ)w JS p, we have that B = R p [τ, ws p ] = R p [τ] + ws p S p. F (τ)b A[τ] + JS p S p = A[τ] + JS p A[τ], since JS R[θ]. Note also that A is integrally closed in B, because R p is integrally closed in S p. The Lemma above now applies, and we can conclude that B = A[τ], and so ws p A[τ]. Since S is module-finite over R[θ], this implies that for some element g R p, gws R[θ]. But g / q and w / q, and so gw / q, while we have just shown that gw J. This is a contradiction, since J q by our choice of q. We next want to define smooth, étale, and unramified algebras. We first need to discuss finitely presented algebras a bit. Let R be any commutative ring. An R-algebra S is called finitely presented if it is finitely generated and for some set of R-algebra generators s 1,..., s n of S, the ideal of polynomial relations on s 1,..., s n over R is a finitely generated ideal. In more detail, note that a choice of R-algebra generators s 1,..., s n yields a homomorphism of a polynomial ring R[X 1,..., X n ] S that is surjective, and so we have that S = R[X 1,..., X n ]/I, where I is the kernel. I is the ideal of polynomial relations on s 1,..., s n over R, and so we are requiring that I have some finite set of generators, say F 1,..., F m. That is, S is finitely presented over R if and only if it has the form R[X 1,..., X n ]/(F 1,..., F m ), where X 1,..., X n are indeterminates. It is reasonable to ask what happens if one chooses a different finite set of algebra generators for S over R. The answer is that the ideal of polynomial relations is still finitely generated. It suffices to compare each set of generators with the union of the two sets, and

13 so we may assume that one set is contained in the other, say s 1,..., s n and s 1,..., s n+h. By induction on h it suffices to consider the case where h = 1, i.e., to compare s 1,..., s n and s 1,..., s n, s where s = s n+1. Since s is in the R-subalgebra generated by s 1,..., s n we can choose a polynomial F R[X 1,..., X n ] such that s = F (s 1,..., s n ). Consider the R-algebra map T = R[X 1,..., X n ] S such that X j s j, 1 j n: let I be the kernel. We extend this to T [X] so that X s. It is easy to verify that the new kernel is J = IT [X] + (X F )T [X]. Clearly, if I is finitely generated, so is J. On the other hand, if J is finitely generated we use the fact that there is an algebra retraction T [X] T that fixes T and maps X to F : the image of J under this map is clearly I, and so a finite set of generators for J will map to a finite set of generators for I. If S is finitely presented over R and T is finitely presented over S then T is finitely presented over R. For suppose that S = R[X 1,..., X n ]/(F 1,..., F m ) and T = S[Y 1,..., Y k ]/(G 1,..., G h ). Each G j can be lifted to an element H j of R[X 1,..., X n, Y 1,..., Y k ] by lifting every coefficient to an element of R[X 1,..., X n ] that maps to it. Then as required. T = R[X 1,..., X n, Y 1,..., Y k ]/(F 1,..., F m, H 1,..., H h ), Finally, note that a localization of R at one (and, hence, at finitely many) elements is finitely presented: R f = R[X]/(fX 1). A localization (at any multiplicative system) of a finitely presented R-algebra is called essentially finitely presented. If the multiplicative system is finitely generated, the word essentially is not needed. Let S be an R-algebra. Let (T, J) be a pair consisting of an R-algebra T and an ideal J T such that J 2 = 0. Then there is an obvious map Θ T,J : Hom R alg (S, T ) Hom R alg (S, T/J) that sends f : S T to its composition γ f with the natural surjection γ : T T/J. By a smooth R-algebra S we mean a finitely presented R-algebra such that for all R-algebras T and J T with J 2 = 0, the map Θ T,J is surjective. By an étale R-algebra S we mean a finitely presented R-algebra such that for all R-algebras T and J T with J 2 = 0, the map Θ T,J is bijective. By an unramified R-algebra S we mean a finitely presented R-algebra such that for all R-algebras T and J T with J 2 = 0, the map Θ T,J is injective. Thus, given an R-algebra map S T/J with J 2 = 0, if S is smooth over R it has at least one lifting to an R-algebra map S T. If S is étale, it has a unique such lifting. If S is unramified it has at most one such lifting (but there may not be any lifting). Note that some authors give these definitions while only requiring S to be essentially finitely presented rather than finitely presented. This creates only small differences in the theory and is convenient in certain ways, while adding a few complications in other ways. 13

14 14 Instead, we shall add the word essentially for this case and talk about maps that are essentially smooth, essentially étale, or essentially unramified. Our next objective is to give many other characterizations of these three properties. These characterizations should offer deep insight into the nature of morphisms with these properties. We first need to review the notions of derivation and of the module of Kähler differentials. A derivation of a ring R into an R-module M is a map D : R M such that for all f, g R, D(f + g) = D(f) + D(g) and D(fg) = fd(g) + gd(f). The kernel of D is a subring of R: note that D(1 1) = 1D(1) + 1D(1) and so D(1) = D(1) + D(1) and D(1) = 0. A derivation D always kills the image of Z in R. If R is an A-algebra then a derivation D : R M is A-linear if and only if the image of A is killed by D. (If it is A-linear then D(a 1) = ad(1) = 0, while if the image of A is killed then D(af) = D ( (a 1)f ) = (a 1)D(f) + fd(a 1) = ad(f) + f 0 = ad(f).) An A-linear derivation is also called an A-derivation. The set of A-derivations Der A (R, M) is an R- submodule of the set of all derivations Der (R, M) of R into M. The module structure may be described as follows: the value of D 1 + D 2 on f R is D 1 (f) + D 2 (f), and the value of rd on f is rd(f). There is a universal A-derivation from R into a specially constructed R-module Ω R/A. We sketch the construction. Let W be the free R-module with a basis in {b f : f R} in bijective correspondence with the elements of R. We want to kill a submodule of W in such a way that the map that takes f R to the image of b f in the quotient of W by this submodule is a derivation. We therefore kill the R-submodule of W spanned by the elements b f+g b f b g and b fg fb g gb f for f, g R, and also b rf rb f for all r R in the image of A and all f R. Let Ω R/A denote the quotient of W by the span V of all these elements. It should be clear that the map d : R Ω R/A that sends f to the image of b f is an A-derivation. We therefore use df to denote the image of b f. The map d : R Ω R/A has the following universal property: given any A-derivation D of R into an R-module M, there is a unique R-linear map T : Ω R/A M such that D = T d. Thus, every A-derivation arises from d, uniquely, by composition with an R-linear map. (It is straightforward to check that the composition of an A-derivation with an R-linear map is an A-derivation.) Otherwise said, for every R-module M we have an isomorphism Hom R (Ω R/A, M) = Der A (R, M). This is the universal mapping property of d : R Ω R/A. (Another way of thinking about this is to note that d : R Ω R/A represents the functor M Der A (R, M) in the category of R-modules.) This mapping property determines d : R Ω R/a uniquely up to unique isomorphism. The proof that every A-derivation D of R M arises uniquely from an R-linear map Ω R/A M is straightforward. Given D, it is clear that if one composes the required linear map with the quotient surjection W Ω R/A, it must map b f D(f) for every f R. This shows uniqueness. There is certainly a unique such map from W to M. All that is needed is to show that it kills all the elements whose span V we took in constructing Ω R/A as W/V. But these are killed precisely because D is an A-derivation. We shall have a lot more to say about derivations and Kähler differentials, but at this point we want to

15 explain how to use them to characterize smooth, étale, and unramified homomorphisms. The proof of the theorem we state next will occupy us for quite a while. Theorem. Let S be a finitely presented R-algebra. (a) If R contains the rationals, S is smooth over R if and only if S is flat over R and Ω S/R is projective as an R-module. (b) S is étale over R if and only if S is flat over R and Ω S/R is 0. (c) S is unramified over R if and only if Ω S/R = 0. In part (a), when R does not necessarily contain the rationals a supplementary condition is needed: e.g., S is smooth over R if and only if for every maximal ideal Q of S lying over P in R, (Ω S/R ) is S Q -free of rank equal to the dimension of S Q /P S Q. Lecture of January 15, 2010 It is obvious from the definition that a homomorphism is étale if and only if it is both smooth and unramified, and that is also obvious from our characterizations using differentials. Note also that the composition of two smooth (respectively, étale, respectively, unramified) homomorphisms is again smooth (respectively, étale, respectively, unramified). For example, suppose that S 2 is smooth over S 1 and S 1 is smooth over R. Given a map S 2 T/J where J 2 = 0 we have a composite map S 1 S 2 T/J which lifts to a map S 1 T. Now, since T is an S 1 -algebra and S 2 is smooth over S 1 we can lift to a map S 2 T. Likewise, given two R-algebra liftings of an R-algebra map S 2 T/J to two maps S 2 T that are distinct, the induced maps S 1 S 2 T must also be distinct, since S 2 is unramified over S 1. But these both lift the same R-algebra map S 1 T/J, contradicting the fact that S 1 is unramified over R. The corresponding result for étale maps is the consequence of putting these two results together. We want to give a bit more feeling for derivations and modules of differentials. Note that by a straightforward induction, if D is a derivation on R then D(f 1 f 2 f k ) = f 2 f k D(f 1 ) + + f 1 f 2 f k 1 D(f k ). A typical term in the sum is the product of all the f i except f j times D(f j ). An immediate consequence is that D(f k ) = kf k 1 D(f). It then follows that D(f k 1 1 f k h h ) is the sum of the h terms of which a typical term is k j f k 1 1 f k j 1 j f k h h D(f j). It also follows that the value of an A-derivation D on a polynomial in f 1,..., f h with coefficients in A is uniquely determined by the values of D(f 1 ),..., D(f h ). 15 In the case of the a polynomial ring B[x] there is B-derivation (we use the partial x derivative notation because B may be a polynomial ring involving other variables) whose value on F (x) is F (x + ) F (x) =0

16 16 where is a new indeterminate. For b B, x (bxk ) = kbx k 1. The polynomial ring R over A in variables x i (there may be infinitely many) has a derivation for each variable x i : one may think of R as B i [x i ] where B i = A[x j : j i]. x i Then if R is an A-algebra, F A[X 1,..., X k ] and f 1,..., f k R, for any A-derivation D we have that D ( F (f 1,..., f k ) ) k F = (f 1,..., f k )D(f i ) x i=1 i. It follows that if we have elements f j that generate R over A then the elements df j span Ω R/A as an R-module. In particular, if R is a finitely generated A-algebra then Ω R/A is a finitely generated R-module. Given a polynomial ring R in variables x i over A, a derivation D of R into M is uniquely determined by specifying values u i M for the variables x i, and it is straightforward to check that there really is a derivation for each specified set of values {u i } i : it sends F to i F x i u i. This implies that for the polynomial ring R, Ω R/A is the free R-module on the dx i. This is true whether the number of variables is finite or infinite. Note that if we have an A-algebra homomorphism R S and an S-module M there is an R-linear map Der A (S, M) Der A (R, M) (where M is thought of as an R-module via restriction of scalars) that is simply induced by composition with the map R S. This implies that there is an R-linear map Ω R/A Ω S/A sends df (this might more precisely be denoted d R/A f) to df (which might more precisely be denoted d S/A f). Hence, there is an S-linear map S R Ω R/A Ω S/A. Let I be an ideal of the A-algebra R. Given an A-derivation R/I M we may compose to get an A-derivation R R/I M. An A-derivation D : R M arises in this way if and only if M is an (R/I)-module and D kills I. It follows that if we have an A-algebra surjection R S with kernel I, then Ω S/A = S R Ω R/A /Span {df : f I}. In the denominator here, it suffices to kill the R-span of the df j for a set of elements f j that generate I (note that d(rf j ) = rd(f j ) + f j dr, and the second summand is 0 because I kills M). Any A-algebra R may be thought of as a polynomial ring T in variables x i modulo the ideal generated by polynomials F j : both i and j may vary in an infinite set here. It follows that Ω R/A may be thought of as the free R-module on the dx i modulo the images of the elements df j calculated in Ω T/A, i.e., the images of the elements F j i x i dx i. In the case of a finitely presented A-algebra R we may work with finitely many variables x i and finitely many polynomials F j, and then Ω R/A is the cokernel of the matrix ( F j ) x i

17 (each entry is replaced by its image in R). This matrix is called the Jacobian matrix. The Jacobian matrix depends on a choice of generators and relations for R over A, but its cokernel, Ω R/A, does not. If R is an A-algebra, W is a multiplicative system in R, and D : R M is a derivation, then any derivation D : R M induces a unique derivation W 1 D : W 1 R W 1 M whose restriction to R is D. Since w(f/w) = f/1 for f R and w W, the extended derivation, if there is one, D, must satisfy D(f)/1 = D(f) = D ( w(f/w) ) w D(f/w)+(f/w) D(w/1) = w D(f/w)+(f/w) ( D(w)/1 ). We may multiply by 1/w to obtain 17 D(f/w) = D(f) w fd(w) w 2 = wd(f) fd(w) w 2, the usual quotient rule. This proves that there is at most one way to define D. It is straightforward to check that if one takes this as the definition (one must check that if f 1 /w 1 = f 2 /w 2 then one gets the same result from either of these representations) then one does in fact get a derivation that extends D. The remaining details are also straightforward. It then follows easily from the universal mapping properties for the modules of differentials and for localization that Ω W 1 R/A = W 1 Ω R/A. Before proving the theorem we have already stated characterizing smooth, étale, and unramified morphisms in terms of the behavior of differentials, we want to give some further characterizations: the proofs of these results are likewise postponed for a while. We need the notion of a geometrically regular algebra over a field. If K is a field, we say that a Noetherian K-algebra R is geometrically regular if for every finite purely inseparable extension L of K, the ring L K R is regular. This implies that R is regular, and in equal characteristic 0, it is equivalent to the condition that R be regular, since the only purely inseparable extension of K is K. Similarly, if K is algebraically closed or perfect, the condition that a K-algebra be geometrically regular is, again, simply the condition that it be regular. It turns out that if R is geometrically regular over K, then L K R is regular for every finite algebraic extension of K. In this generality, infinite field extensions L are slightly problematic in that one may lose the Noetherian property. However, when R is essentially of finite type over K, i.e., a localization of a finitely generated K-algebra, the geometric regularity of R over K implies that L K R is regular for every field extension L of K. In this case it turns out that R is geometrically regular provided that L R K is regular for some field that contains a perfect closure of K (in characteristic p, this is a maximal purely inseparable algebraic extension, gotten by adjoining all p e th roots of all elements). In particular, if R is essentially of finite type over K, then R is geometrically regular if and only if K K R is regular, where K is an algebraic closure of K.